From: Metroplex Math Circle To: bhardwajbhavana@yahoo.com Sent: Friday, October 7, 2016 11:12 AM Subject: [New post] October 10, 2016 – AMC 8 Registration #yiv4096759756 a:hover {color:red;}#yiv4096759756 a {text-decoration:none;color:#0088cc;}#yiv4096759756 a.yiv4096759756primaryactionlink:link, #yiv4096759756 a.yiv4096759756primaryactionlink:visited {background-color:#2585B2;color:#fff;}#yiv4096759756 a.yiv4096759756primaryactionlink:hover, #yiv4096759756 a.yiv4096759756primaryactionlink:active {background-color:#11729E;color:#fff;}#yiv4096759756 WordPress.com | Metroplex Math Circle posted: “For those of you who were unable to attend the last math circle and, therefore, also unable to sign up for the AMC 8, a registration will be held on October 10th at UTD from 4-6pm at the Visitor’s Center (look for the table across from the Bookstore). T” | |

]]>It is the best book for practice

But I m not able to deal with few ques.

So I want solutions for them so that they may become a helpful source beside my teacher….

Plz help

I need solutions…for XII class

Please send them via mail to me in .pdf format

Thank you… ]]>

[(1 +1/r)^n – (1 -1/r)^n]^1/n = 2^1/n [(n*1)1/r + (n*3) 1/r^3 …]^1/n.

Let r =1, so that p=q. This deliberately converts the 3 term assumption of FLT into 2 terms. Hence when r =1 and p=q, there are just 2 terms,

[(n*1) + (n*3)…..]^1/n = 2^(1- 1/n).

It also follows that provided n is greater than 2 the Pythagorean triples power, and provided the 3 term assumption is restored, [(n*1) 1/r + (n*3)1/r^3 …]^1/n will be less than 2 when multiplied by by 2^1/n. there can only be equality with 2 terms not the the 3 terms assumed in Fermat’s Last Theorem. The Wiles Proof of Fermat’s Last Theorem lacks any explanation of the absence of the power n=2 being the Pythagorean triples situation from the conclusions of FLT. This could mean that the whole proof is wrong particularly since it is only thought to be understood by a small clique of mathematicians. ]]>